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(i) CH2Cl2
(ii) CHCl3
(iii) CCl4 Asked by Veena Last Modified 
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Answer
Vishaka Bhutia
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(i)
Dichlormethane (CH2Cl2)
μ = 1.60D
(ii)
Chloroform (CHCl3)
μ = 1.08D
(iii)
Carbon tetrachloride (CCl4)
μ = 0D
CCl4 is a symmetrical molecule. Therefore, the dipole moments of all four C−Cl bonds cancel each other. Hence, its resultant dipole moment is zero.
As shown in the above figure, in CHCl3, the resultant of dipole moments of two C−Cl bonds is opposed by the resultant of dipole moments of one C−H bond and one C−Cl bond. Since the resultant of one C−H bond and one C−Cl bond dipole moments is smaller than two C−Cl bonds, the opposition is to a small extent. As a result, CHCl3 has a small dipole moment of 1.08 D.
On the other hand, in case of CH2Cl2, the resultant of the dipole moments of two C−Cl bonds is strengthened by the resultant of the dipole moments of two C−H bonds. As a result, CH2Cl2 has a higher dipole moment of 1.60 D than CHCl3 i.e., CH2Cl2 has the highest dipole moment.
Hence, the given compounds can be arranged in the increasing order of their dipole moments as:
CCl4 < CHCl3 < CH2Cl2
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